Question

The value of the expression ${{}^{n+1}C}_2+2[{{}^{2}C}_2+{{}^{3}C}_2+{{}^{4}C}_2+......+{{}^{n}C}_2]$ is $\frac{n(n+k)(pn+m)}{h}$. Find $k+m+p+h$ ?

Answer 1

${}^2{C}_2={}^3{C}_3$
Hence
${}^2{C}_2+{}^3{C}_2+{}^4{C}_2+...{}^n{C}_2$
$={}^3{C}_3+{}^3{C}_2+{}^4{C}_2+...{}^n{C}_2$
$={}^4{C}_3+{}^4{C}_2+{}^5{C}_2+...{}^n{C}_2$ ...$({}^n{C}_r+{}^n{C}_{r+1}={}^{n+1}{C}_{r+1})$.
:
:
${}^n{C}_3+{}^n{C}_2$
$={}^{n+1}{C}_3$
Hence
${}^{n+1}{C}_2+2{}^{n+1}{C}_3$
$={}^{n+1}{C}_2+{}^{n+1}{C}_3+{}^{n+1}{C}_3$
$={}^{n+2}{C}_3+{}^{n+1}{C}_3$
$=\frac{n(n+1)(n+2)}{6}+\frac{(n-1)n(n+1)}{6}$
$=\frac{\left(n\right)(n+1)[n+2+n-1]}{6}$
$=\frac{n(n+1)(2n+1)}{6}$
Therefore $k=1$, $m=1$, $p=2$ and $h=6$
Hence
$k+m+p+h$
$=10$.